For a radial function in three dimensions, the radial Laplacian gives

\nabla^{2} ϕ = \frac{1}{r ^{2}} (r^{2} ϕ^{'})^{'} .

(1)

Here regularity at zero selects

r^{2} ϕ^{'} (r) = \int_{0}^{r} s^{2} (s - 1) e^{s} d s = e^{r} (r^{3} - 4 r^{2} + 8 r - 8) + 8.

(2)

A further integration gives

ϕ (r) = \frac{e ^{r} ( r ^{2} - 5 r + 8 ) - 8}{r} + C,

(3)

whose apparent singularity is removable, with limit

3 + C

at zero. Since the nonconstant term has value

4 e - 8

r = 1

, the boundary condition

ϕ (1) = 3

gives

C = 11 - 4 e

. Therefore

ϕ (r) = \frac{e ^{r} ( r ^{2} - 5 r + 8 ) - 8}{r} + 11 - 4 e .

(4)

ii

 0  0

Solution

 0  0

The radial equation is

(r^{2} ϕ^{'})^{'} = - \frac{1}{r} .

(1)

Its solutions satisfying decay at infinity have the form

ϕ (r) = \frac{l o g r + 1 + C}{r} .

(2)

The boundary value

ϕ (1) = 1

gives

C = 0

, so

ϕ (r) = \frac{1 + lo g r}{r} .

(3)

It tends to zero as

r \to \infty

and directly satisfies

\nabla^{2} ϕ = - r^{- 3}

iii

 0  0

Solution

 0  0

Regularity at zero gives

r^{2} ϕ^{'} (r) = \int_{0}^{r} \frac{s ^{2}}{1 + s ^{2}}, d s = r - arctan r .

(1)

Integrating once more,

ϕ (r) = \frac{1}{2} lo g (1 + r^{2}) + \frac{a r c t a n r}{r} + C,

(2)

where

arctan r / r

has removable limit one at zero. At

r = 1

, the nonconstant terms sum to

\frac{1}{2} lo g 2 + \frac{π}{4}

. Hence

ϕ (r) = \frac{1}{2} lo g (1 + r^{2}) + \frac{arctan r}{r} - 1 - \frac{1}{2} lo g 2 - \frac{π}{4} .

(3)

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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 9B / c

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